[tex]\textit{volume of a hemisphere}\\\\ V=\cfrac{1}{2}\cdot \cfrac{4\pi r^3}{3}~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ V=74466 \end{cases}\implies 74466=\cfrac{1}{2}\cdot \cfrac{4\pi r^3}{3} \\\\\\ 74466=\cfrac{2\pi r^3}{3}\implies 223398=2\pi r^3\implies \cfrac{22398}{2\pi }=r^3\implies \cfrac{111699}{\pi }=r^3 \\\\\\ \sqrt[3]{\cfrac{111699}{\pi }}=r~\hfill \stackrel{\textit{diameter = 2r}}{2\sqrt[3]{\cfrac{111699}{\pi }}\implies \sqrt[3]{\cfrac{893592}{\pi }}}~~ \boxed{\approx~~65.77}[/tex]
Evaluate the triple integral. X dv, where e is bounded by the paraboloid x = 4y2 4z2 and the plane x = 4. E
The triple integral that is bounded by a paraboloid x = 4y2 4z2 given as 16.762
Parabloid, x = 4y² + 4z²
plane x = 4
x = 4y² + 4z²
x = 4
4 = 4y² + 4z²
4 = 4 (y² + z² )
1 = y² + z²
from polar coordinates
y = r cos θ
z = r sin θ
r² = y² + z²
The limits of the integral0 ≤ θ ≤ 2π
4r² ≤ x ≤ 4
0 ≤ r ≤ 1
[tex]\int\limits\int\limits\int\limits {x} \, dV = \int\limits^1_0\int\limits^a_b\int\limits^c_d {x} \, dx ( rdrdz)[/tex]
where
a = 4
b = 4r²
c = 2r
d = 0
The first integral using limits c and d gives:
[tex]2pi\int\limits^1_0\int\limits^a_b {xr} \, dx[/tex]
The second integral using limits a and b
[tex]pi\int\limits^1_0 {16 } } \, rdr - pi\int\limits^1_0 {16r^{5} \, dx[/tex]
[tex]16pi\int\limits^1_0 { } } \, rdr - 16pi\int\limits^1_0 {r^{5} \, dx[/tex]
[tex]16pi\int\limits^1_0 { } } \, [r-r^{5}]dr[/tex]
The third integral using limits 1 and 0 gives: 16.762
Read more on Triple integral here: https://brainly.com/question/27171802
The triple integral that is bounded by a paraboloid x = 4y2 4z2 given as 16.762
What is integration?Integration is defined as adding small parts to form a new significant part.
Parabloid, x = 4y² + 4z²
plane x = 4
x = 4y² + 4z²
x = 4
4 = 4y² + 4z²
4 = 4 (y² + z² )
1 = y² + z²
from polar coordinates
y = r cos θ
z = r sin θ
r² = y² + z²
The limits of the integral
0 ≤ θ ≤ 2π
4r² ≤ x ≤ 4
0 ≤ r ≤ 1
[tex]\int\int\intxdV = \int_0_1\int_b_a\int_d_cxdx(rdrdz)[/tex]
where
a = 4
b = 4r²
c = 2r
d = 0
The first integral using limits c and d gives:
[tex]2\pi\int_0^1\int_b^axydx[/tex]
The second integral using limits a and b
[tex]\pi \int_0^116rdr-\pi\int_0^116r^5dx[/tex]
[tex]16\pi \int_0^1rdr-16\pi\int_0^1 r^5dx[/tex]
[tex]16\pi\int_0^1[r-r^5]dr[/tex]
The third integral using limits 1 and 0 gives: 16.762
Read more on Triple integral here:
brainly.com/question/27171802
#SPJ4
A cyclist travels a distance of 837 1/2 feet in 25 seconds. The cyclist travels at a constant rate. What is the unit rate, in feet per second, at which the cyclist travels?
PLEASE
Answer:
[tex]\boxed{33.5 \ \text{feet/second}}[/tex]
Step-by-step explanation:
To find the unit rate, we need to find out how much distance does the cyclist travel in 1 second. For that, we need to use the unitary method. A unitary method is a method that determines the unit rate of an object.
Note: Since the cyclist travels at a constant rate, unitary method can be used.
[tex]\rightarrow 837 \dfrac{1}{2} \ \text{feet in 25 seconds}[/tex]
[tex]\rightarrow 837.5 \ \text{feet} = 25 \ \text{seconds}[/tex]
Divide both sides by 25:
[tex]\rightarrow \dfrac{837.5}{25} \ \text{feet} = \dfrac{25}{25} \ \text{seconds}[/tex]
[tex]\rightarrow 33.5 \ \text{feet} = 1 \ \text{seconds}[/tex]
[tex]\rightarrow \boxed{33.5 \ \text{feet/second}}[/tex]
Learn more about unitary method: https://brainly.com/question/19423643
4x+y=2 solve for y algebra 2
Answer:
Step-by-step explanation:
4x + y = 2
y = -4x + 2
Please help! I don't know what answer b is, but i know a, thank you.
They are corresponding angles
Step-by-step explanation:
We know that x=75⁰ because
[tex]180 - 105 = 75[/tex]
75⁰ is angle EFB
Since we know this, and because we know that AD//EH,
We can conclude that x=75⁰(AD is parallel to EH and angle CBA or x and angle EFB are corresponding angles)
[tex] \sf \large \: find \: \frac{dy}{dx} \: if \: x=a ( \Theta + sin \Theta ) , y= a (1- Cos \Theta ) \: at \: \Theta = \frac{π}{2} \\ \\ \\ \\ \\ [/tex]
Kindly Don't Spàm!
Thanks!!!!
Given :
[tex] \: \: \: [/tex]
[tex] \rm \large \: x = a ( \Theta + Sin \Theta )[/tex][tex] \: \: \: [/tex]
[tex] \rm \large y = a ( 1 - cos \: \Theta )[/tex][tex] \: \: [/tex]
Now , x = a ( θ + sin θ )
[tex] \: \: [/tex]
Diff w.r.t " θ "
[tex] \: \: \: [/tex]
[tex] \rm \large\frac{dx}{dθ } = a \: \frac{d}{dθ} (θ + \sin\theta )[/tex][tex] \: \: [/tex]
[tex] \boxed{ \rm \large\underline{ \frac{dx}{d \theta} = a(1 + \cos \theta ) }}[/tex][tex] \: \: [/tex]
Now y = a ( 1-cosθ)
[tex] \: \: [/tex]
Diff w.r.t " θ " we get .
[tex] \: \: [/tex]
[tex] \rm \large \frac{dy}{d \theta} = a \frac{d}{d \theta} (1 - cos \theta) [/tex][tex] \: \: \: [/tex]
[tex] \boxed{ \rm \large \underline{ \frac{dy}{d \theta} = a \sin \theta}}[/tex][tex] \: \: \: [/tex]
From eqn ( 1 ) & ( 2 )
[tex] \: \: [/tex]
[tex] \rm \large \frac{dy}{dx} = \frac{ \frac{dy}{d \theta} }{\frac{dx}{d \theta}} [/tex][tex] \: \: \: [/tex]
[tex] \: \: \: \rm \large = \frac{ \cancel{a} \: sin \theta}{ \cancel a \: (1 + cos \theta)} [/tex][tex] \: \: [/tex]
[tex] \rm \large \: = \frac{sin \theta}{1 + \cos \theta } [/tex][tex] \: \: \: [/tex]
[tex] \rm \large \: \frac{2 \sin( \frac{\theta }{2} ) cos\frac{\theta }{2} }{2 \: cos ^{2} \frac{\theta }{2}} \: \: ......(sin \: a \: = 2 \: sin \frac{a}{2} \: cos \frac{a}{2} 1 + \: cos \: a \: = 2cos ^{2} \frac{a}{2} )[/tex][tex] \: \: [/tex]
[tex] \rm \large \: \frac{dy}{dx} = \frac{sin \frac{ \theta}{2} }{cos \frac{ \theta}{2} } [/tex][tex] \: \: [/tex]
[tex] \rm \large \: \frac{dy}{dx} = tan \frac{ \theta}{2} [/tex][tex] \: \: [/tex]
[tex] \rm \large \: ( \frac{dy}{dx} ) = tan \frac{ \frac{ \theta}{2} }{2} [/tex][tex] \: \: [/tex]
[tex] \rm \large \: = tan( \frac{ \theta}{4} )[/tex][tex] \: \: [/tex]
[tex] \boxed{ \rm \large \underline{ ( \frac{dy}{dx} ) = \frac{\pi}{2} = 1}}[/tex][tex] \: \: [/tex]
Hope Helps!:)